StrmCkr wrote:take this current hardest puzzle - try solving it with out using a guess and test method ps if you do succumb to trial and error this is a back door size 3 puzzle which requires 3 different clues added to the puzzle at the same time to reach a single solution.

Thanks for underlining my point that using pencil marks important, if not outright necessary.

eleven wrote:Can you solve this one with pencilmarks ? It is easy without

I have no idea how to solve it without pm, but with them it wasn't very hard. Such a solution is obviously off-topic here, but mine is below anyway. Wouldn't mind if you enlightened us with the no-pm way!

The puzzle is by ssxsssxs, the technique by qiuyanzhe. It is a very nice extension (one of three) of RW's reverse BUG lite.In columns 5 and 6 (same stack) you can see a triple 124, with only one number in different rows. So it has distinction 1.Rows 2 and 3 (same band) have the same numbers 3 an 4. If 1 or 5 would be in r2c1 (can't be in r2c8), you would have a triple too, with one extra digit in r3, it would have distinction 1 too.But this is not possible, if the puzzle is unique. Therefore 1 must be in r2c3, and the puzzle is solved very easy.

And yes, it is very rare, that you can use that technique, but so cool.

Thanks, eleven! Can't say that I fully understand the technique, but cool it is! I just checked that Hodoku's default solution for this puzzle takes a whopping 23 steps, so it's a pretty effective step indeed (and quite a bit more elegant than Hodoku's one-step net option). I doubt I'll be adding it to my solving arsenal any time soon, though.

I have noticed the existence of this topic for long but have never read through the text. Today is my first time browsing them, and as a competitive sudoku player, I have my things to share. There is so much thing so I would just tell a little bit now.

In competition any marking(or try&error, even try-without-error) is allowed, and the only criterium is the solving time. Personally my act is similar to RW's, but I would make some Pencilmarks, but limited to Double candidates, or cells with "unordinary" pencilmark deletions. I learned from grandmasters to mark intersections/pairs between cells, and summarized what deleting a candidate can make. Marking bivalue cells could help discovering (Naked Pairs), (Naked Singles after Intersection/..), (XY-Wing-like chains), or (ab-bc-ca triplets)..

Also I would keep the pencilmarks in bivalue cells ordered. If we have 12 23 13in a minirow, I would mark 12 23 31,such that two possibility of all these cells have two possibilities-- all left or all right. If any of their columns has the same property, then ordered bivalue can spread and eliminate some candidates.This is like branching, and I think it suits competing.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~And I wonder what this topic should be about.When I first see this topic, instead of "How to solve without marking on the grid", I thought this topic is about "How to discover(and use) informations other than pencilmarks" as pencilmark is not the only form of information. I think this is more open than what we discussed in the beginning.

In 999_Spring's post, the 67-UR becomes the bottleneck of some of the puzzles, while I immediately see that UR is useless. There is a Sue De Coq, which is also what I saw.SE:UR(with hidden triplet, 4.7, eliminate 3@r5c5) SE sees the UR as a non-candidate information"r56c6 cannot both be 67", while r3c6's 67 directly shows the same. But it can only utilize such kind of information directly in a UR. And if we change the 67 pair to multivalue "half-ur"(like a 67-75-56) SE uses forcing chains. This confuses me as two problems, one is "why only two-value URs count", and the other is "Why only UR-form non-candidate information can be utilized".Below is the puzzle (Simplified the UR part).

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~During years of my sudoku career I have used non-candidate methods uncountable number of times, while the first time I discover it was when drawing 6*6 jigsaw sudoku empty grids. I found that some shapes can directly determine equalities.

And the first time I used equation in an actual puzzle is like this(when I was 10 and haven't heard of this forum): Original Puzzle:...........3.....1.8..9..7......6....7....8.....1.3.9.........679..5........4...3 (edit:I cannot find the source anyway.. Bought a book and didn't find the puzzle)

Above is the bottleneck of a 17-cell puzzle, by a UR we can get an 8 in r7c6 but it doesn't break the puzzle. Its bottleneck is a chain on 4(SE7.7).And here's what I solved it:According to r5 and c5, (suppose)a=45=r4c4=r5c9, then r3c3=a(hidden single in row), and r9c2=a(hidden single in column), a≠4, a=5.

Such ways can be used in multiple ways, it is hard to tell with a fixed pattern, as another example in a PK is:

r9c1=39, hidden single in b4 we have r5c2=r9c1=39, then we get an intersection on 2, so r6c6=7.

It also works on some 999999111/ 999991110-like puzzles. Partial grids solvable by this have various SE ratings up to 8.9 ...~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~Still a bit too long.. For things not about specific grids I would post to Coffee Bar: What I See Here before the new year.